10. A flowerpot is dropped from the balcony of an apartment,
28.5 m above the ground. At a time of 1.00 s after the pot is
dropped, a ball is thrown vertically downward from the bal-
cony one storey below, 26.0 m above the ground. The initial
velocity of the ball is 12.0 m/s [down]. Does the ball pass
the flowerpot before striking the ground? If so, how far
above the ground are the two objects when the ball passes
the flowerpot?
To find out if the ball passes the flowerpot before striking the ground, we first need to calculate the time it takes for the ball to reach the ground. We can use the following kinematic equation for the ball's vertical motion:
d = v_i * t + 0.5 * a * t^2
Where:
d = distance traveled
v_i = initial velocity
t = time elapsed
a = acceleration (in this case, due to gravity, approximated as -9.8 m/s^2)
Solving for t:
0 = 12.0 * t - 0.5 * 9.8 * t^2
Rearranging the equation:
4.9t^2 - 12t = 0
Factoring out t:
t(4.9t - 12) = 0
So, either t = 0 (initial time) or 4.9t - 12 = 0
Let's solve for t using the second equation:
4.9t - 12 = 0
4.9t = 12
t = 12/4.9
t ≈ 2.45 seconds
Now, let's calculate how far above the ground the ball and the flowerpot are at t = 1.00 second.
For the ball:
d_ball = v_i * t + 0.5 * a * t^2
d_ball = 12.0 * 1.00 + 0.5 * 9.8 * 1.00^2
d_ball = 12.0 + 0.5 * 9.8 * 1.00
d_ball = 12.0 + 4.9
d_ball ≈ 16.9 meters
For the flowerpot:
d_pot = v_i * t + 0.5 * a * t^2
d_pot = 0 * 1.00 + 0.5 * 9.8 * 1.00^2
d_pot = 0 + 0.5 * 9.8 * 1.00
d_pot = 0 + 4.9
d_pot = 4.9 meters
Therefore, when t = 1.00 second, the ball is about 16.9 meters above the ground, and the flowerpot is about 4.9 meters above the ground. So, the ball passes the flowerpot before striking the ground.